Let\({v_1} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{4}}\end{array}} \right]\),\({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{0}}\end{array}} \right]\), and let\(S = \left\{ {{v_1},{v_2},{v_3}} \right\}\).

1. Show that the set is affinely independent.
2. Find the barycentric coordinates of\({p_1} = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\),\({p_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\end{array}} \right]\),\({p_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{\bf{1}}\end{array}} \right]\),\({p_{\bf{4}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\end{array}} \right]\), and\({p_{\bf{5}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{1}}\end{array}} \right]\), with respect to S.
3. Let\(T\)be the triangle with vertices\({v_1}\),\({v_{\bf{2}}}\), and\({v_{\bf{3}}}\). When the sides of\(T\)are extended, the lines divide\({\mathbb{R}^{\bf{2}}}\)into seven regions. See Figure 8. Note the signs of the barycentric coordinates of the points in each region. For example,\({{\bf{p}}_{\bf{5}}}\)is inside the triangle\(T\)and all its barycentric coordinates are positive. Point\({{\bf{p}}_{\bf{1}}}\)has coordinates\(\left( { - , + , + } \right)\). Its third coordinate is positive because\({{\bf{p}}_{\bf{1}}}\)is on the\({{\bf{v}}_{\bf{3}}}\)side of the line through\({{\bf{v}}_{\bf{1}}}\)and\({{\bf{v}}_{\bf{2}}}\). Its first coordinate is negative because\({{\bf{p}}_{\bf{1}}}\)is opposite the\({{\bf{v}}_{\bf{1}}}\)side of the line through\({{\bf{v}}_{\bf{2}}}\)and\({{\bf{v}}_{\bf{3}}}\). Point\({{\bf{p}}_{\bf{2}}}\)is on the\({{\bf{v}}_{\bf{2}}}{{\bf{v}}_{\bf{3}}}\)edge of\(T\). Its coordinates are\(\left( {0, + , + } \right)\). Without calculating the actual values, determine the signs of the barycentric coordinates of points\({{\bf{p}}_{\bf{6}}}\),\({{\bf{p}}_{\bf{7}}}\), and\({{\bf{p}}_{\bf{8}}}\)as shown in Figure 8.

The set is said to be affinely dependent if the set\(\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},...,{{\bf{v}}_p}} \right\}\)in the dimension\({\mathbb{R}^n}\)exists such that for non-zero scalars\({c_1},{c_2},...,{c_p}\), the sum of scalars is zero i.e.\({c_1} + {c_2} + ... + {c_p} = 0\), and\({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_p}{{\bf{v}}_p} = 0\).

(a)

Consider the set of vectors\(S = \left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},{{\bf{v}}_3}} \right\}\), where\({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\),\({{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}0\\4\end{array}} \right]\), and\({{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}2\\0\end{array}} \right]\).

Let the newpoints, \({{\bf{v}}_3} - {{\bf{v}}_1}\) and \({{\bf{v}}_2} - {{\bf{v}}_1}\), be obtained by eliminating the first point.

\(\begin{array}{c}{{\bf{v}}_3} - {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}2\\0\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}3\\{ - 2}\end{array}} \right]\end{array}\)

And

\(\begin{array}{c}{{\bf{v}}_2} - {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}0\\4\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\end{array}\)

It is observed that \({{\bf{v}}_2} - {{\bf{v}}_1}\) and \({{\bf{v}}_3} - {{\bf{v}}_1}\) are not proportional or multiples of each other. They are linearly independent. Also, the sum of weights cannot be 0.

Thus, the indexed set \(S\)is affinely independent.

(b)

Obtain the barycentric coordinates of points using the vectors\({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\),\({{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}0\\4\end{array}} \right]\),\({{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}2\\0\end{array}} \right]\), and the points\({{\bf{p}}_1} = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\),\({{\bf{p}}_2} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\),\({{\bf{p}}_3} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right]\),\[{{\bf{p}}_4} = \left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]\], and\({{\bf{p}}_5} = \left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\)as shown below:

Construct an augmented matrix using the above vectors and points.

\(M = \left[ {\begin{array}{*{20}{c}}{ - 1}&0&2&2&1&{ - 2}&1&1\\2&4&0&3&2&1&{ - 1}&1\\1&1&1&1&1&1&1&1\end{array}} \right]\)

Obtain the row-reduced echelon form as shown below:

Interchange rows 1 and 2.

\(M = \left[ {\begin{array}{*{20}{c}}2&4&0&3&2&1&{ - 1}&1\\{ - 1}&0&2&2&1&{ - 2}&1&1\\1&1&1&1&1&1&1&1\end{array}} \right]\)

Multiply row 1 by\(\frac{1}{2}\)to get row 1.

\(M = \left[ {\begin{array}{*{20}{c}}1&2&0&{3/2}&1&{1/2}&{ - 1/2}&{1/2}\\{ - 1}&0&2&2&1&{ - 2}&1&1\\1&1&1&1&1&1&1&1\end{array}} \right]\)

Add rows 1 and 2 to get row 2. Then, add\( - 1\)times row 1 to row 3 to get row 3.

\(M = \left[ {\begin{array}{*{20}{c}}1&2&0&{3/2}&1&{1/2}&{ - 1/2}&{1/2}\\0&1&1&{7/4}&1&{ - 3/4}&{1/4}&{3/4}\\0&0&2&{5/4}&1&{ - 1/4}&{7/4}&{5/4}\end{array}} \right]\)

Multiply row 3 by\(\frac{1}{2}\)to get row 3.

\(M = \left[ {\begin{array}{*{20}{c}}1&2&0&{3/2}&1&{1/2}&{ - 1/2}&{1/2}\\0&1&1&{7/4}&1&{ - 3/4}&{1/4}&{3/4}\\0&0&1&{5/8}&{1/2}&{ - 1/8}&{7/8}&{5/8}\end{array}} \right]\)

Add\( - 1\)times row 3 to row 2 to get row 2. Then, add\( - 2\)times row 2 to row 1 to get row 1.

\(M = \left[ {\begin{array}{*{20}{c}}1&0&0&{ - 6/8}&0&{14/8}&{6/8}&{1/4}\\0&1&0&{9/8}&{1/2}&{ - 5/8}&{ - 5/8}&{1/8}\\0&0&1&{5/8}&{1/2}&{ - 1/8}&{7/8}&{5/8}\end{array}} \right]\)

Thus, the barycentric coordinates of points are\({{\bf{p}}_1} \leftrightarrow \left( { - \frac{6}{8},\frac{9}{8},\frac{5}{8}} \right)\),\({{\bf{p}}_2} \leftrightarrow \left( {0,\frac{1}{2},\frac{1}{2}} \right)\), \({{\bf{p}}_3} \leftrightarrow \left( {\frac{{14}}{8}, - \frac{5}{8}, - \frac{1}{8}} \right)\), \({{\bf{p}}_4} \leftrightarrow \left( {\frac{6}{8}, - \frac{5}{8},\frac{7}{8}} \right)\), and \({{\bf{p}}_5} \leftrightarrow \left( {\frac{1}{4},\frac{1}{8},\frac{5}{8}} \right)\).

(c)

To obtain the signs of the coordinates \({{\bf{p}}_6}\), \({{\bf{p}}_7}\), and \({{\bf{p}}_8}\), observe the given figure:

Point\({{\bf{p}}_6}\)should have the coordinates\(\left( { - , - , + } \right)\)because it lies outside the triangle. Also, the third coordinate is positive because\({{\bf{p}}_6}\)is on the\({{\bf{v}}_3}\)side of the line through vectors\({{\bf{v}}_1}\)and\({{\bf{v}}_2}\).

Point\({{\bf{p}}_7}\)should have the coordinates\(\left( {0, + , - } \right)\)because it lies outside the triangle. Also, the second coordinate is positive because\({{\bf{p}}_7}\)is on the\({{\bf{v}}_1}\)side of the line through vectors\({{\bf{v}}_2}\)and\({{\bf{v}}_3}\).

Point\({{\bf{p}}_8}\)should have the coordinates\(\left( { + , + , - } \right)\)because it lies outside the triangle. Also, the first and second coordinates are positive because\({{\bf{p}}_8}\)is on the\({{\bf{v}}_1}\)side of the line through vectors\({{\bf{v}}_2}\)and\({{\bf{v}}_3}\).

Therefore, the signs of the barycentric points are \({{\bf{p}}_6} \leftrightarrow \left( { - , - , + } \right)\), \({{\bf{p}}_7} \leftrightarrow \left( {0, + , - } \right)\), and \({{\bf{p}}_8} \leftrightarrow \left( { + , + , - } \right)\).